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Q: Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to \bigtriangleup ABC with scale factor \frac{3}{2} . Justify the construction. Are the two triangles congruent? Note that all the three angles and two sides of the two triangles are equal

Solution

Steps of construction

  1. Draw a line segment BC = 6 cm
  2. Taking B and C as a centre, draw an arc of radius AB= 4cm and AC=9cm
  3. Join AB and AC
  4. DABC is required triangle From B draw ray BM with acute angle \angle XBM
  5. Make 3 points B_{1},B_{2},B_{3}  on BM with equal distance
  6. Join B_{2}C  and B_{3}  draw B_{3}X\parallel B_{2}C  intersecting BC at X From point X draw XY||CA intersecting the extended line segment BA to Y Then \bigtriangleup BXY  is the required triangle whose sides are equal to\frac{3}{2}  of the \bigtriangleup ABC
    Justification :
    Here B_{3}X\parallel B_{2}C
    \therefore \frac{BC}{CX}= \frac{2}{1}
    \therefore \frac{BX}{BC}= \frac{BC+CX}{BC}= 1+\frac{1}{2}= \frac{3}{2}
    Also XY\parallel CA
    \bigtriangleup ABC\sim \bigtriangleup YBX
    \therefore \frac{YB}{AB}= \frac{YX}{AC}= \frac{BX}{BC}= \frac{3}{2}
    Here all three angles are the same but the three sides are not the same.
    \thereforeThe two triangles are not congruent because, if two triangles are congruent, then they have the same shape and same size.

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Q: Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is $60^{\circ}$. Also, justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.

Solution

Steps of construction:
1. Construct a circle with $O$ as a centre and a 4 cm radius
2. Construct any diameter $A O B$
3. Construct an angle $\angle A O P=600$ where $O P$ is the radius which intersect the circle at the point $P$
4. Construct $P Q$ perpendicular to $O P$ and $B E$ perpendicular to $O B$
$P Q$ and $B E$ intersect at the point $R$
5. $R P$ and $R B$ are the required tangents
6. The measurement of $O R$ is 8 cm

Justification:
PR is the tangent to a circle

$$
\angle O P Q=900
$$

$B R$ is the tangent to a circle

$$
\angle O B R=900
$$
So we get

$$
\angle P O B=180-60=1200
$$
In BOPR

$$
\angle B R P=360-(120+90+90)=600
$$
Therefore, the distance between the centre of the circle and the point of intersection of tangents is 8 cm.

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Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ABC= 60^{\circ}. Construct a triangle similar to ABC with a scale factor \frac{5}{7} . Justify the construction.

Solution
Given : AB = 5 cm, BC = 6 cm

Steps of construction

  1. Draw a line segment AB = 5 cm
  2. draw < ABO= 60^{\circ}  B taking as a centre draw an arc of radius BC=6cm
  3. Join AC, DABC is the required triangle
  4.  From point A draw any ray A{A}'   with an acute angle \angle BA{A}'
  5.  Mark 7 points B_{1},B_{2},B_{3},B_{4},B_{5},B_{6},B_{7}  with equal distance.
  6. Join B_{7}B  and form B_{5}  draw B_{5}X\parallel B_{7}B\, \, BYmaking the angle equal From point X draw XY\parallel BC intersecting AC at Y. Then, DAMN is the required triangle whose sides are equal to \frac{5}{7}  the corresponding sides of the \bigtriangleup ABC.

Justification: Here, B_{5X}\parallel B_{7}B  [by construction]
          \therefore \frac{AX}{XB}= \frac{5}{2}\Rightarrow \frac{XB}{AX}= \frac{2}{5}
Now \frac{AB}{AX}= \frac{AX+XB}{AX}
1+\frac{XB}{AX}= 1+\frac{2}{5}= \frac{7}{5}
Also, XY\parallel BC
\therefore \bigtriangleup AXY\sim \bigtriangleup ABC
\frac{AX}{AB}= \frac{AY}{AC}= \frac{YX}{BC}= \frac{5}{7}

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Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.

Solution

Steps of construction
1. Draw line $B C=5 \mathrm{~cm}$
2. Taking $B$ and $C$ as centres, draw two arcs of equal radius 6 cm intersecting each other at point $A$.
3. Join $A B$ and $A C D A B C$ is required isosceles triangle

4 From B draw ray $B_X$ with an acute angle $C B B^{\prime}$
6. draw $B_1, B_2, B_3, B_4$ at $B X$ with equal distance
7. Join $B_3 C$ and from $B_4$ the draw line $B_4 D \| B_3 C$, intersect extended segment BC at point D .

Then EBD is required triangle. We can name it PQR.

$$
\begin{aligned}
& \text { Justification } \\
& \because B_4 D \| B_3 C \\
& \therefore \frac{B C}{C D}=\frac{3}{1} \Rightarrow \frac{C D}{B C}=\frac{1}{3} \\
& \text { Now } \therefore \frac{B D}{B C}=\frac{B C+C D}{B C}=1+\frac{C D}{B C}=1+\frac{1}{3}=\frac{4}{3} \\
& \text { Also } D E \| C A \\
& \therefore \triangle A B C \sim \triangle D B E \\
& \frac{E B}{A B}=\frac{D E}{C A}=\frac{B D}{B C}=\frac{4}{3}
\end{aligned}
$$

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Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on the outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.

Solution
 

Steps of construction
1.   Draw two concentric circles with centre O and radii 3 cm and 5 cm
2.   Taking any point P on the outer circle, Join P and O
3.   Draw a perpendicular bisector of OP let M be the midpoint of OP
4.   Taking M as the centre and OM as the radius draw a circle which cuts the inner circle at Q and R
5.   Join PQ and PR. Thus PQ and PR are required tangents 
On measuring PQ and PR we find that PQ = PR = 4 cm
Calculations
\bigtriangleup OQP, \angle OQP= 90^{\circ}
OP^{2}= OQ^{2}+PQ^{2} [using Pythagoras theorem]
\left ( 5 \right )^{2}= \left ( 3 \right )^{2}+\left ( PQ \right )^{2}
25-9= PQ^{2}
16= PQ^{2}
\sqrt{16}= PQ
4cm= PQ
Hence the length of both tangents is 4 cm.

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Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and ABC= 60^{\circ}, divide it into triangles BCD and ABD
by the diagonal BD. Construct the triangle BD'C' similar to DBDC with scale factor  \frac{4}{3}. Draw the line segment D'A' parallel to DA where A' lies on the extended side BA. Is A'BC'D' a parallelogram?

Solution

Steps of construction
1.   Draw, A-line AB = 3 cm
2  Draw a ray by making  \angle ABP= 60^{\circ}
3.   Taking  B as the centre and radius equal to 5 cm. Draw an arc which cuts BP at point C
4.   Again draw ray AX making  \angle {Q}'AX= 60^{\circ}
5.   With A as the centre and radius equal to 5 cm draw an arc which cuts AX at point D
6.   Join C and D Here ABCD is a parallelogram           
7.   Join BD , BD is a diagonal of parallelogram ABCD
8.   From B draw a ray BQ with any acute angle at point B i.e., \angle CBQ  is an acute angle
9.   Locate 4 points B_{1},B_{2},B_{3},B_{4}  on BQ with equal distance.
10. Join B_{3}C  and from  B_{4},{C}'  parallel to B_{3}C  which intersect at point {C}'
 11. From point  {C}' draw a line {C}'{D}'  which is parallel to CD
 12. Now draw a line segment {D}'{A}'  parallel to the DA
 Note: Here {A}',{C}'  and {D}'  are the extended sides.
13.  {A}'B{C}'{D}' is a parallelogram in which {A}'{D}'= 6\cdot 5\, cm  and {A}'{B}= 4\, cm  and < {A}'B{D}'= 60^{\circ} divide it into triangles B{C}'{D}'  and {A}'{BD}'  by the diagonal {BD}'

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Two line segments AB and AC include an angle of 60^{\circ} where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC, respectively such that AP= \frac{3}{4}AB  and AQ= \frac{1}{4}AC . Join P and Q and measure the length PQ.

Solution
Given  :
AB = 5 cm and AC = 7 cm
AP= \frac{3}{4}AB\: \: \cdots 1
AQ= \frac{1}{4}AC\: \: \cdots 2
From equation 1
AP= \frac{3}{4}AB
AP= \frac{3}{4}\times 5= \frac{15}{4}\: \: \left [ \because AB= 5cm \right ]
P is any point on B
\therefore PB= AB-AP= 5-\frac{15}{4}= \frac{20-15}{4}= \frac{5}{4}cm
\frac{AP}{AB}= \frac{15}{4}\times \frac{4}{5}= \frac{1}{3}
AP:AB= 1:3
\thereforescale of a line segment AB is \frac{1}{3}


 Steps of construction
  1.   Draw a line segment AB = 5 cm
  2.   Now draw ray AO which makes an angle i.e., < BAO= 60^{\circ}
  3.   Which A as centre and radius equal to 7 cm draw an arc cutting line AO at C
  4.   Draw ray AP with acute angle BAP
  5.   Along AP make 4 points A_{1},A_{2},A_{3},A_{4}  with equal distance.
  6.   Join A_{4}B
  7.   From  A_{3} draw A_{3}P  which is parallel to A_{4}B  which meet AB at point P.
Then P is a point which divides AB in a ratio 3 : 1
   AP : PB = 3 : 1
  8.   Now draw ray AQ, with an acute angle CAQ.
  9.   Along AQ mark 4 points  B_{1},B_{2},B_{3},B_{4} with equal distance.
 10. Join  B_{4}C
  11. From B_{1}  draw B_{1}Q  which is parallel to B_{4}C  which meet AC at point Q.
  Then Q is a point which divides AC in a ratio 1 : 3
   AQ : QC = 1 : 3
  12. Finally join PQ and its measurement is 3.25 cm.

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Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre.

Solution 
Given : Radius = 4cm

Steps of construction
1.  Draw a circle at radius r = 4 cm at point O. 
2.   Take a point P at a distance of 6cm from point O and join PO
3.  Draw a perpendicular bisector of line PO M is the mid-point of  PO.
4.   Taking M as the centre draw another circle of radius equal to MO and it intersects the given circle at points Q and R  

5.   Now, join PQ and PR.
Then PQ and PR one the required two tangents.

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Draw a triangle ABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm. Construct a triangle similar to it and of scale factor. \frac{5}{3}

Solution 
Given :  BC = 6cm, CA = 5cm, AB = 4cm
Scale factor = 5/3
Let m = 5 and n = 3

Steps of construction

  1. Draw line BC = 6cm
  2. Taking B and C as centre mark arcs of length 4cm and 5cm respectively which intersect each other at point A.
  3. Join BA and CA
  4. Draw a ray BX making an acute angle with BC.
  5. Locate 5 Points on BX in equidistant as B1, B2, B3, B4, B5.
  6. Join B3 to C and draw a line through B5 parallel to B3C to intersect at BC extended at  {C}'.

       7. Draw a line through  {C}' parallel to AC intersect AB extended at {A}' .

Now  {A}'B{C}'  is the required triangle.

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Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and \angle B= 90^{\circ}. Construct a triangle similar to it and of scale factor \frac{2}{3} . Is the new triangle also a right triangle?

Solution
  Given BC = 12cm, AB = 5 cm, \angle B= 90^{\circ}
 

Steps of construction
1.   Draw a line BC = 12 cm
2.   From B draw AB = 5 cm which makes an angle of 90^{\circ} at B.
3.   Join AC
4.   Make an acute angle at B as < CBX
5.   On BX mark 3 points at equal distance X_{1},X_{2},X_{3} .
6.   Join X_{3}C
7.   From X_{2}  draw X_{2}{C}'\parallel X_{3}C  intersect AB at {C}'
8.   From point {C}'  draw {C}'{A}'\parallel CA  intersect AB {A}'
Now  \bigtriangleup {A}'B{C}' is the required triangle and  \bigtriangleup {A}'B{C}'  is also a right triangle.

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